Translating Graphs: Shifts of a Parent Function

Start from a parent function like \(f(x) = x^2\) and you can slide its whole graph anywhere on the plane without changing its shape at all — add a constant outside for a vertical move, subtract one inside for a horizontal move.

By the end you'll be able to read \(f(x-h)+k\) and say exactly how far, and which way, the graph of f moved — including catching the horizontal sign flip that trips most people up.

Predict: does \(f(x-3)\) shift the graph of \(f(x)=x^2\) left or right? Set h = 3 below and check.

This plots \(g(x) = f(x-h) + k\) for the parent \(f(x)=x^2\), shown as the faint dotted ghost. Drag h to move the graph horizontally and k to move it vertically — the vertex, still sitting at (0, 0) on the ghost, rides along to \((h, k)\). Hover the curve or focus the vertex dot for exact readouts.

g(x) = f(x − 0.0) + 0.0 = (x − 0.0)² + 0.0; vertex (0.0, 0.0) — no shift yet.

g(x) = f(x−h)+k — translated curve, vertex, and the parent \(f(x)=x^2\) as a ghost
g(x) = f(x−h)+k f(x) = x² (parent) h shift k shift

A translation slides a graph across the plane without touching its shape — \(f(x)+k\) moves it up or down, and \(f(x-h)\) moves it left or right.

Visual

Picture the parent \(f(x)=x^2\) sitting with its vertex at the origin. Add \(k\) outside the function and the whole curve rides an elevator — straight up for \(k>0\), straight down for \(k<0\). Subtract \(h\) inside the parentheses and the curve slides sideways instead — right for \(h>0\), left for \(h<0\). Neither move bends, stretches, or flips the shape; it's the same U, just relocated.

Formal

For any function \(f\) and constants \(h, k\):

  • Vertical shift: \(f(x) + k\) moves every output by \(k\) — up when \(k>0\), down when \(k<0\). This one behaves exactly like it looks.
  • Horizontal shift: \(f(x-h)\) moves every input requirement by \(h\) — right when \(h>0\), left when \(h<0\). This one runs backwards from what the sign suggests: to reach the output the parent had at some \(x\), the new graph needs an input that is \(h\) larger, so the whole curve slides right by \(h\). Reading \(x-h\) as "the graph moves toward \(+h\)" — so \(x-3\) moves toward \(+3\), i.e. right — helps it stick.

Combine both and \(g(x) = f(x-h)+k\) moves the reference point (like a parabola's vertex) from \((0,0)\) to \((h,k)\) exactly.

Applied

If a physics model for an object's height is \(f(t) = -16t^2\) (dropped at \(t=0\) from height 0), then \(f(t-2)+5\) describes the same fall started 2 seconds later and from 5 feet higher up — the shape of the fall never changes, only when and where it starts.

Worked example

Graph \(g(x) = (x-3)^2 + 2\) starting from \(f(x)=x^2\). Identify the pieces: since it's written \(x-3\), \(h=3\); and \(k=2\). Horizontal shift: right 3, moving the vertex from (0, 0) to (3, 0). Vertical shift: up 2, moving it to (3, 2). Check a point: the parent passes through (1, 1); after shifting right 3 and up 2 it should pass through (4, 3): \(g(4) = (4-3)^2+2 = 1+2 = 3\). It matches.

Your turn

Same parent, different numbers: \(g(x) = (x+2)^2 + 4\). Rewrite \(x+2\) as \(x-(-2)\), so \(h = \) ____ and \(k = 4\). That means the vertex moves from (0, 0) to ____.

Reveal the answer

\(h = -2\) (left 2, not right 2 — the plus sign is the trap), and \(k=4\) (up 4). Vertex moves to (-2, 4). Set h = -2, k = 4 on the sliders above and check the vertex label against this answer.

More info — why f(x−h) shifts right, not left

It helps to track a single output. The parent \(f(x)=x^2\) hits its vertex value 0 at \(x=0\). For \(g(x)=f(x-h)\) to hit that same value 0, you need \(x-h=0\), i.e. \(x=h\). So the point that used to sit at \(x=0\) on the parent now sits at \(x=h\) on \(g\) — every point of the graph has moved \(h\) units to the right when \(h>0\). The same logic in reverse explains why \(f(x+h)\) (which is \(f(x-(-h))\)) shifts left. See the Paul's Online Notes link in Dive deeper for more worked examples of this reasoning.

Check your understanding

Question 1 of 4

Describe the graph of \(g(x) = (x+4)^2 - 1\) relative to the parent \(f(x) = x^2\).

Question 2 of 4

You want the vertex of \(f(x) = x^2\) to move from (0, 0) to (-6, 0) — a horizontal move only. Which function achieves this?

Question 3 of 4

For \(f(x) = x^2\), compare \(g(x) = f(x) + 3\) and \(h(x) = f(x-3)\). Which correctly describes both shifts?

Question 4 of 4

Let \(f(x) = 2x - 1\) (function notation from an earlier lesson). Define \(g(x) = f(x-3) + 2\). Find \(g(5)\).

Recap

  • A translation slides a graph without changing its shape — never a rotation, stretch, or flip.
  • Vertical shift: \(f(x)+k\) moves the graph up \(k\) units for \(k>0\), down for \(k<0\) — behaves exactly like it looks.
  • Horizontal shift: \(f(x-h)\) moves the graph right \(h\) units for \(h>0\), left for \(h<0\) — the sign runs backwards from intuition, so double-check it.
  • Combined, \(g(x)=f(x-h)+k\) moves a reference point like a vertex from (0, 0) to \((h,k)\).

Dive deeper

Sources

  • Translating Graphs (Vertical and Horizontal Shifts)